Pythonic way to combine for-loop and if-statement

You can use generator expressions like this:

gen = (x for x in xyz if x not in a)

for x in gen:
    print(x)

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As per The Zen of Python (if you are wondering whether your code is "Pythonic", that's the place to go):

Beautiful is better than ugly.

Explicit is better than implicit.

Simple is better than complex.

Flat is better than nested.

Readability counts.

The Pythonic way of getting the sorted intersection of two sets is:

>>> sorted(set(a).intersection(xyz))
[0, 4, 6, 7, 9]

Or those elements that are xyz but not in a:

>>> sorted(set(xyz).difference(a))
[12, 242]

But for a more complicated loop you may want to flatten it by iterating over a well-named generator expression and/or calling out to a well-named function. Trying to fit everything on one line is rarely "Pythonic".

Update following additional comments on your question and the accepted answer

I'm not sure what you are trying to do with enumerate, but if a is a dictionary, you probably want to use the keys, like this:

>>> a = {
...     2: 'Turtle Doves',
...     3: 'French Hens',
...     4: 'Colly Birds',
...     5: 'Gold Rings',
...     6: 'Geese-a-Laying',
...     7: 'Swans-a-Swimming',
...     8: 'Maids-a-Milking',
...     9: 'Ladies Dancing',
...     0: 'Camel Books',
... }
>>>
>>> xyz = [0, 12, 4, 6, 242, 7, 9]
>>>
>>> known_things = sorted(set(a.iterkeys()).intersection(xyz))
>>> unknown_things = sorted(set(xyz).difference(a.iterkeys()))
>>>
>>> for thing in known_things:
...     print 'I know about', a[thing]
...
I know about Camel Books
I know about Colly Birds
I know about Geese-a-Laying
I know about Swans-a-Swimming
I know about Ladies Dancing
>>> print '...but...'
...but...
>>>
>>> for thing in unknown_things:
...     print "I don't know what happened on the {0}th day of Christmas".format(thing)
...
I don't know what happened on the 12th day of Christmas
I don't know what happened on the 242th day of Christmas

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The following is a simplification/one liner from the accepted answer:

a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]

for x in (x for x in xyz if x not in a):
    print(x)

12
242

Notice that the generator was kept inline. This was tested on python2.7 and python3.6 (notice the parens in the print ;) )

It is honestly cumbersome even so: the x is mentioned four times.

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I personally think this is the prettiest version:

a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]
for x in filter(lambda w: w in a, xyz):
  print x

Edit

if you are very keen on avoiding to use lambda you can use partial function application and use the operator module (that provides functions of most operators).

https://docs.python.org/2/library/operator.html#module-operator

from operator import contains
from functools import partial
print(list(filter(partial(contains, a), xyz)))

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I would probably use:

for x in xyz: 
    if x not in a:
        print(x...)

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a = [2,3,4,5,6,7,8,9,0]
xyz = [0,12,4,6,242,7,9]  
set(a) & set(xyz)  
set([0, 9, 4, 6, 7])

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You can use generators too, if generator expressions become too involved or complex:

def gen():
    for x in xyz:
        if x in a:
            yield x

for x in gen():
    print x

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I liked Alex's answer, because a filter is exactly an if applied to a list, so if you want to explore a subset of a list given a condition, this seems to be the most natural way

mylist = [1,2,3,4,5]
another_list = [2,3,4]

wanted = lambda x:x in another_list

for x in filter(wanted, mylist):
    print(x)

this method is useful for the separation of concerns, if the condition function changes, the only code to fiddle with is the function itself

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

for x in filter(wanted, mylist):
    print(x)

The generator method seems better when you don't want members of the list, but a modification of said members, which seems more fit to a generator

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

generator = (x**0.5 for x in mylist if wanted(x))

for x in generator:
    print(x)

Also, filters work with generators, although in this case it isn't efficient

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

generator = (x**0.9 for x in mylist)

for x in filter(wanted, generator):
    print(x)

But of course, it would still be nice to write like this:

mylist = [1,2,3,4,5]

wanted = lambda x:(x**0.5) > 10**0.3

# for x in filter(wanted, mylist):
for x in mylist if wanted(x):
    print(x)

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Use intersection or intersection_update

intersection : a = [2,3,4,5,6,7,8,9,0] xyz = [0,12,4,6,242,7,9] ans = sorted(set(a).intersection(set(xyz)))

intersection_update: a = [2,3,4,5,6,7,8,9,0] xyz = [0,12,4,6,242,7,9] b = set(a) b.intersection_update(xyz) then b is your answer

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A simple way to find unique common elements of lists a and b:

a = [1,2,3]
b = [3,6,2]
for both in set(a) & set(b):
    print(both)

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based on the article here: https://towardsdatascience.com/a-comprehensive-hands-on-guide-to-transfer-learning-with-real-world-applications-in-deep-learning-212bf3b2f27a I used the following code for the same reason and it worked just fine:

an_array = [x for x in xyz if x not in a]

This line is a part of the program! this means that XYZ is an array which is to be defined and assigned previously, and also the variable a

Using generator expressions (which is recommended in the selected answer) makes some difficulties because the result is not an array

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